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Kuliah 1 Mobile Computing. Basis teori Komunikasi Data. Analisa Fourier Bandwidth-Limited Signals Maximum Data Rate dari Channel. Analisa Fourier. ∞. ∞. n=1. n=1. T. 0. T. 0. T. 0. Setiap suatu fungsi periodik g(t) dengan periode T dapat dinyatakan sebagai :

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kuliah 1 mobile computing
Kuliah 1

Mobile Computing

basis teori komunikasi data
Basis teori Komunikasi Data
  • Analisa Fourier
  • Bandwidth-Limited Signals
  • Maximum Data Rate dari Channel
analisa fourier
Analisa Fourier









Setiapsuatufungsiperiodik g(t) denganperiode T dapatdinyatakansebagai:

g(t) = c/2 + ∑ an sin(2∏nft) + ∑ bncos(2∏nft)

Dimana: an = 2/T ∫ g(t) sin(2∏nft) dt

bn = 2/T ∫ g(t) cos(2∏nft) dt

cn = 2/T ∫ g(t) dt

bandwidth limited signals
Bandwidth-Limited Signals

A binary signal and its root-mean-square Fourier amplitudes.

(b) – (c) Successive approximations to the original signal.

bandwidth limited signals 2
Bandwidth-Limited Signals (2)

(d) – (e) Successive approximations to the original signal.

hubungan antara bandwidth dengan data rate
Hubungan antara bandwidth dengan data rate
  • Kasus 1: AndaikansinyalpadaGambar 2a didekatidenganGambar 2e (8 harmonisa), dan f = 1 MHz; maka bandwidth = (8-1) x1 MHz = 7 MHz, dan data rate = ?? Mbps
  • Kasus 2 AndaikansinyalpadaGambar 2a didekatidenganGambar 2e (8 harmonisa), dan bandwidth 15 MHz; maka data rate = ?? Mbps
  • Kasus 3: AndaikansinyalpadaGambar 2a didekatidenganGambar 2d (4 harmonisa), dan bandwidth 7 MHz; maka data rate = ?? Mbps
  • Kesimpulan ???
kapasitas dari kanal channel
Kapasitas dari kanal (channel)
  • Nyquist:

Maximum data rate (C) = 2B log2 M

  • Shannon:

Maximum data rate (C) = B log2 (1+S/N)

C : kapasitas kanal (bps)

B : bandwidth (Hz)

M : jumlah level tegangan

S/N : signal to noise ratio

dalam decibel: S/N = 10 log10 S/N

  • Andaikan spektrum suatu kanal berada diantara 3 MHz dan 4 MHz, dan S/N = 24 dB, maka

B = 4 MHz – 3 MHz = 1 Mhz

S/N = 24 dB = 10 log10 S/N  S/N = 251

  • Dengan formula Shannon:

C = 106 log2 (1+251) ≈ 106 x 8 = 8 Mbps

  • Dengan formula Nyquist:

C = 2B log2 M

8 x 106 = 2 x 106 x log2 M

4 = log2 M  M = 16


(a) A binary signal

(b) Amplitude modulation

(c) Frequency modulation

(d) Phase modulation

modem dengan m berbeda
Modem dengan M berbeda

(a) QPSK (Quadrature Phase Shift Keying)

(b) QAM-16 (Quadrature Amplitude Modulation -16)

(c) QAM-64 (Quadrature Amplitude Modulation - 64)

wireless transmission
Wireless Transmission
  • The Electromagnetic Spectrum
  • Radio Transmission
  • Microwave Transmission
  • Infrared and Millimeter Waves
  • Lightwave Transmission
the electromagnetic spectrum
The Electromagnetic Spectrum

The electromagnetic spectrum and its uses for communication.

radio transmission
Radio Transmission

(a) In the VLF, LF, and MF bands, radio waves follow the curvature of the earth.

(b) In the HF band, they bounce off the ionosphere.

lightwave transmission
Lightwave Transmission

Convection currents can interfere with laser communication systems.

A bidirectional system with two lasers is pictured here.

fiber optics
Fiber Optics

(a) Three examples of a light ray from inside a silica fiber impinging on the air/silica boundary at different angles.

(b) Light trapped by total internal reflection.

transmission of light through fiber
Transmission of Light through Fiber

Attenuation of light through fiber in the infrared region.

fiber cables
Fiber Cables

(a) Side view of a single fiber.

(b) End view of a sheath with three fibers.

  • Frequency Division Multiplexing
  • Wavelength Division Multiplexing
  • Time Division Multiplexing
  • Code Division Multiple Access
frequency division multiplexing
Frequency Division Multiplexing

(a) The original bandwidths.

(b) The bandwidths raised in frequency.

(b) The multiplexed channel.

wavelength division multiplexing
Wavelength Division Multiplexing

Wavelength division multiplexing.

time division multiplexing
Time Division Multiplexing

The T1 carrier (1.544 Mbps).

time division multiplexing 2
Time Division Multiplexing (2)

Multiplexing T1 streams into higher carriers.

cdma code division multiple access
CDMA – Code Division Multiple Access

(a) Binary chip sequences for four stations

(b) Bipolar chip sequences

(c) Six examples of transmissions

(d) Recovery of station C’s signal

  • Circuit Switching
  • Message Switching
  • Packet Switching
circuit switching
Circuit Switching

(a) Circuit switching.

(b) Packet switching.

message switching
Message Switching

(a) Circuit switching (b) Message switching (c) Packet switching

packet switching
Packet Switching

A comparison of circuit switched and packet-switched networks.

the mobile telephone system
The Mobile Telephone System
  • First-Generation Mobile Phones: Analog Voice
  • Second-Generation Mobile Phones: Digital Voice
  • Third-Generation Mobile Phones:Digital Voice and Data
advanced mobile phone system
Advanced Mobile Phone System

(a) Frequencies are not reused in adjacent cells.

(b) To add more users, smaller cells can be used.

gsm global system for mobile communications
GSMGlobal System for Mobile Communications

GSM uses 124 frequency channels, each of which uses an eight-slot TDM system

gsm 2
GSM (2)

A portion of the GSM framing structure.

third generation mobile phones digital voice and data
Third-Generation Mobile Phones:Digital Voice and Data

Basic services an IMT-2000 network should provide

  • High-quality voice transmission
  • Messaging (replace e-mail, fax, SMS, chat, etc.)
  • Multimedia (music, videos, films, TV, etc.)
  • Internet access (web surfing, w/multimedia.)